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## Homework Statement

An infinite wire of linear charge density [itex]\lambda[/itex] lies on the [itex]z[/itex] axis. An insulating cylindrical shell of radius [itex]R[/itex] is concentric with the wire and can rotate freely about the [itex]z[/itex] axis. The charge per unit area on the cylinder is [itex]\sigma = -\lambda/2\pi R[/itex] while the mass per unit area is [itex]\rho[/itex]. A magnetic field [itex]\mathbf{B} = B\hat{z}[/itex] fills the region of the cylinder.

At [itex]t=0[/itex] the cylinder is at rest and the magnetic field is reduced such that [itex]\mathbf{B}(t) = Be^{-t/\tau}\hat{z}[/itex].

Use conservation of angular momentum to find the angular velocity [itex]\omega[/itex] of the shell for [itex]t\to\infty[/itex].

## Homework Equations

Maxwell.

## The Attempt at a Solution

Inside the region of the shell, the magnetic field is initially just [itex]\mathbf{B} = B\hat{z}[/itex]. The electric field from a wire is [itex]\mathbf{E} = \lambda \hat{s}/2\pi \epsilon_{0} s[/itex]. Therefore, the momentum density is [itex]\mathbf{p} = \epsilon_{0}\mathbf{E}\times\mathbf{B} = \lambda B \hat{\phi}/2\pi s[/itex]. The angular momentum density is [itex]\mathbf{l} = \mathbf{s}\times \mathbf{p} = \lambda \hat{z}B/2\pi[/itex]. The total angular momentum per length is therefore [itex]\mathbf{L} = \int \mathbf{l} \,\text{d}V = \frac{1}{2}\lambda B R^2 \hat{z}[/itex].

Now, at [itex]t \to \infty[/itex], the ambient magnetic field goes to zero, so the angular momentum remaining is the angular momentum due to the spinning cylinder plus the angular momentum due to the magnetic field it's making and the electric field from the wire. When the cylinder is rotating, it's basically a solenoid, so the standard Amperian loop of hright [itex]h[/itex] gives [itex]B_{z}h = \mu_{0} I = \mu_{0} \frac{\text{d}Q}{\text{d} t} = \mu_{0} \frac{\text{d}Q}{\text{d} A}\frac{\text{d}A}{\text{d} t} = \mu_{0} \sigma \frac{\text{d}A}{\text{d} t} [/itex]. But then, [itex]\text{d} A = h \text{d} s = hR\text{d}\theta = hR\omega \text{d} t\implies \frac{\text{d}A}{\text{d} t} = h R \omega[/itex] so [itex]\mathbf{B} = \mu_{0}\sigma R \omega \hat{z} = -\frac{\mu_{0}}{2\pi}\lambda \omega \hat{z} [/itex]. This is just [itex]\mathbf{B} = B' \hat{z}[/itex] for [itex]B' = -\frac{\mu_{0}}{2\pi}\lambda \omega[/itex], so the result for the angular momentum of this field with the electric field still holds, and [itex]\mathbf{L}_{\text{field}} =\frac{1}{2}\lambda B' R^2 \hat{z} [/itex]. Finally, via conservation of momentum, [itex]\mathbf{L}_{0} = \mathbf{L}_{\text{field}} + \mathbf{L}_{\text{spin}}\implies \frac{1}{2}\lambda B R^2 = \frac{1}{2}\lambda B' R^2 + I \omega[/itex]. Also, the moment of inertia per length is [itex]I = mR^2 = \rho \frac{A}{h} R^2 = 2\pi \rho R^4[/itex], so we have [itex]\omega = 2\pi B\lambda/(8\pi^2 \rho - \mu_{0}\lambda^2)[/itex].

The issue is that I get I much simpler result via Faraday's Law, and since it was a lot simpler I assumed that this is somehow wrong.

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